Find the domain of the function f(x) = log_{[ | vedclass

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(B) For the function $f(x)$ to be defined, the following conditions must be satisfied:$1$. For binomial coefficients and permutations, the parameters

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$\{4, 5\}$

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(B) For the function $f(x)$ to be defined, the following conditions must be satisfied:$1$. For binomial coefficients and permutations, the parameters must be non-negative integers:For $^{16-x}C_{2x-1}$, we need $16-x \ge 2x-1 \ge 0$, which implies $3x \le 17$ $(x \le 5.66)$ and $2x \ge 1$ $(x \ge 0.5)$.For $^{20-3x}P_{2x-5}$, we need $20-3x \ge 2x-5 \ge 0$, which implies $5x \le 25$ $(x \le 5)$ and $2x \ge 5$ $(x \ge 2.5)$.Combining these, we get $2.5 \le x \le 5$. Since $n$ and $r$ in $^nC_r$ and $^nP_r$ must be integers, $2x-1$ and $2x-5$ must be integers, implying $2x$ must be an integer.$2$. The base of the logarithm $[x + \frac{1}{x}]$ must be $> 0$ and $
eq 1$. For $x \in [2.5, 5]$, $x + \frac{1}{x} \ge 2.5 + 0.4 = 2.9$, so $[x + \frac{1}{x}] \ge 2$. This condition is satisfied.$3$. The argument $|x^2 - x - 6| > 0$ implies $x^2 - x - 6 
eq 0$, so $x 
eq 3$ and $x 
eq -2$.Testing integer values for $2x$ in $[2.5, 5]$: $x \in \{2.5, 3, 3.5, 4, 4.5, 5\}$.If $x=3$, $|3^2-3-6| = 0$, which is undefined.If $x=4$, $f(4) = \log_{[4.25]} |6| + ^{12}C_7 + ^{8}P_3$, which is defined.If $x=5$, $f(5) = \log_{[5.2]} |14| + ^{11}C_9 + ^{5}P_5$, which is defined.If $x=2.5$, $2x-1=4$ and $2x-5=0$, $f(2.5) = \log_{[2.9]} |6.25-2.5-6| + ^{13.5}C_4$ (undefined as $n$ must be integer).Thus, the domain is $\{4, 5\}$.

Find the domain of the function $f(x) = \log_{[x + \frac{1}{x}]} |x^2 - x - 6| + ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$, where $[x]$ denotes the greatest integer function.

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